A crate is pulled to the right with a force
of 93.5 N, to the left with a force of 116.8 N,
upward with a force of 658.1 N, and downward
with a force of 248.6 N.
What is the net external force in the x
direction?
Answer in units of N
To find the net external force in the x direction, we need to calculate the vector sum of the forces in the x direction.
Given:
Force to the right = 93.5 N
Force to the left = 116.8 N
Since the force to the right is positive and the force to the left is negative, we can simply subtract the force to the left from the force to the right to find the net external force in the x direction:
Net external force in the x direction = Force to the right - Force to the left
= 93.5 N - 116.8 N
= -23.3 N
Therefore, the net external force in the x direction is -23.3 N.
To find the net external force in the x direction, we need to consider the forces acting in that direction. In this case, the crate is being pulled to the right with a force of 93.5 N and to the left with a force of 116.8 N.
To determine the net external force in the x direction, we can find the difference between these two forces:
Net external force in the x direction = Force to the right - Force to the left
Net external force in the x direction = 93.5 N - 116.8 N
Net external force in the x direction = -23.3 N
Therefore, the net external force in the x direction is -23.3 N. It has a negative sign because the force to the left is greater than the force to the right.
X = -116.8 + 93.5 = -23.3 N.
Y = -248.6 + 658.1 = 409.5 N.
Tan Ar = Y/X = 409.5/-23.3 = -17.57511
Ar = -86.7o = Reference angle.
A = -86.7 + 180 = 93.3o, CCW.
Fr = Y/sin A = 409.5/sin93.3 = 410 N.
[93.3o]. = Resultant force.
Fx = 410*Cos93.3 = -23.61 N = 23.61 N.,
Left(West).